Problem: As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as <image>, where k is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.

A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number n, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!

Input

The first input line contains an integer n (1 ≤ n ≤ 109).

Output

Print "YES" (without the quotes), if n can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).

Examples

Input

256


Output

YES


Input

512


Output

NO

Note

In the first sample number <image>.

In the second sample number 512 can not be represented as a sum of two triangular numbers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
int getbit(int i, T X) {
  return ((X >> (i - 1)) & 1);
}
template <class T>
T onbit(int i, T X) {
  return (X | (1 << (i - 1)));
}
template <class T>
T offbit(int i, T X) {
  return ((X | (1 << (i - 1))) - (1 << (i - 1)));
}
template <class T>
T sqr(T x) {
  return (x * x);
}
template <class T>
T cube(T x) {
  return (x * x * x);
}
template <class T>
T gcd(T a, T b) {
  T r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
int csx[4] = {0, 0, -1, 1};
int csy[4] = {-1, 1, 0, 0};
long long n;
long long find(long long K) {
  long long L, R;
  L = 1;
  R = K;
  while (L <= R) {
    long long mid = (L + R) >> 1;
    long long cur = mid * (mid + 1) / 2;
    if (cur == K) return mid;
    if (cur > K) R = mid - 1;
    if (cur < K) L = mid + 1;
  }
  return -1;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin >> n;
  for (typeof((long long)(sqrt)(n)) i = 1; i <= (long long)(sqrt)(n); i++) {
    long long P = n - i * (i + 1) / 2;
    if (find(P) != -1) {
      cout << "YES";
      return 0;
    }
  }
  cout << "NO";
  return 0;
}